2Classical gases

II Statistical Physics

2.5 Interacting gases

So far, we have been talking about ideal gases. What happens if there are

interactions?

For a real gas, if they are sufficiently dilute, i.e.

N/V

is small, then we expect

the interactions to be negligible. This suggests that we can try to capture the

effects of interactions perturbatively in

N

V

. We can write the ideal gas law as

p

kT

=

N

V

.

We can think of this as a first term in an expansion, and add higher order terms

p

kT

=

N

V

+ B

2

(T )

N

2

V

2

+ B

3

(T )

N

3

V

3

+ ··· .

Note that the coefficients depend on

T

only, as they should be intensive quantities.

This is called the Virial expansion, and the coefficients

B

k

(

T

) are the Virial

coefficients. Our goal is to figure out what these B

k

(T ) are.

We suppose the interaction is given by a potential energy

U

(

r

) between two

neutral atoms (assuming monoatomic) at separation r.

Example. In genuine atoms, for large r (relative to atomic size), we have

U(r) ∝ −

1

r

6

.

This comes from dipole-dipole interactions. Heuristically, we can understand

this power of 6 as follows — while the expectation values of electric dipole of an

atom vanishes, there exists non-trivial probability that the dipole

p

1

is non-zero.

This gives an electric field of

E ∼

p

1

r

3

.

This induces a dipole p

2

in atom 2. So we have

p

2

∝ E ∼

p

1

r

3

.

So the resulting potential energy is

U ∝ −p

2

E ∼ −

p

2

1

r

6

.

This is called the van der Waals interaction. Note that the negative sign means

this is an attractive force.

For small

r

, the electron orbitals of the atoms start to overlap, and then we get

repulsion due to the Pauli principle. All together, we obtain the Lennard-Jones

potential given by

U(r) = U

0

r

0

r

12

−

r

0

r

6

.

r

U(r)

To make life easy, we will actually use a “hard core repulsion” potential instead,

given by

U(r) =

(

∞ r < r

0

−U

0

r

0

r

6

r > r

0

r

U(r)

r

0

For a general U, we can write the Hamiltonian of the gas is

H =

N

X

i=1

p

2

i

2m

+

X

i>j

U(r

ij

),

with

r

ij

= |r

i

− r

j

|.

is the separation between particle i and particle j.

Because of this interaction, it is no longer the case that the partition function

for

N

particles is the

N

th power of the partition function for one particle. We

have

Z(N, V, T ) =

1

N

1

(2π~)

3n

Z

N

Y

i=1

d

3

p

i

d

3

r

i

e

−βH

=

1

N!

1

(2π~)

3n

Z

Y

i

d

3

p

i

e

−βp

2

i

/2m

!

Z

Y

i

d

3

r

i

e

−β

P

j<k

U(r

jk

)

!

=

1

N!λ

3N

Z

Y

i

d

3

r

i

e

−β

P

j<k

U(r

jk

)

,

where again

λ =

r

2π~

2

mkT

.

Since we want to get an expansion, we might want to expand this in terms of

the potential, but that is not helpful, because the potential is infinite for small

r

.

Instead it is useful to consider the Mayer f-function:

f(r) = e

−βU(r)

− 1.

This function has the property that

f

(

r

) =

−

1 for

r < r

0

(in the case of the

hardcore repulsion), and

f

(

r

)

→

0 as

r → ∞

. So this is a nicer function as it

only varies within this finite range.

We further simplify notation by defining

f

ij

= f(r

ij

).

Then we have

Z(N, V, T ) =

1

N!λ

3N

Z

Y

i

d

3

r

i

Y

j<k

(1 + f

jk

)

=

1

N!λ

3N

Z

Y

i

d

3

r

i

1 +

X

j<k

f

jk

+

X

j<k

X

`<m

f

jk

f

`m

+ ···

.

The first term is just

Z

Y

i

d

3

r

i

= V

N

,

and this gives the ideal gas term. Now each of the second terms is the same, e.g.

for j = 1, k = 2, this is

Z

Y

i

d

3

r

i

f

12

= V

N−2

Z

d

3

r

1

d

3

r

2

f

12

= V

N−1

I,

where we set

I =

Z

d

3

r f (r).

Since

f

(

r

)

→

0 as

r → ∞

, we might as well integrate over all space. Summing

over all terms, and approximating

N

(

N −

1)

/

2

∼ N

2

/

2, we find that the first

two terms of the partition function are

Z(N, V, T ) =

V

N

N!λ

3N

1 +

N

2

2V

I + ···

.

Up to first order, we can write this as

Z(N, V, T ) =

V

N

N!λ

3N

1 +

N

2V

I + ···

N

= Z

ideal

1 +

N

2V

I + ···

N

.

This pulling out of the

N

to the exponent might seem a bit arbitrary, but writing

it this way, it makes it much clearer that

S

,

F

etc would be extensive quantities.

For example, we can write down the free energy as

F = −kT log Z = F

ideal

− NkT log

1 +

N

2V

I + ···

.

Without actually computing

I

, we can expect that it grows as

I ∼ r

3

0

. Since we

are expanding in terms of NI/V , we need

N

V

1

r

3

0

.

So we know the expansion is valid if the density of the gas is much less than the

density of an atom. In real life, to determine the density of an atom, we need to

find a system where the atoms are closely packed. Thus, it suffices to measure

the density of the substance in liquid or solid form.

Assuming that the gas is indeed not dense, we can further use the approxi-

mation

log(1 + x) ≈ x.

So we have

p = −

∂F

∂V

T

=

NkT

V

1 −

N

2V

I + ···

.

So we have

pV

NkT

= 1 −

N

2V

I + ··· .

So we can now read off what the second Virial coefficient is:

B

2

= −

1

2

I.

We can consider what we get with different potentials. If we have a completely

repulsive potential, then

U

(

r

)

>

0 everywhere. So

f <

0, and thus

B

2

(

t

)

>

0. In

other words, having a repulsive interaction tends to increase the pressure, which

makes sense. On the other hand, if we have an attractive potential, then the

pressure decreases.

If we have a hardcore repulsion, then we have

I =

Z

r=r

0

r=0

d

3

r (−1) +

Z

∞

r=r

0

d

3

r

e

βU

0

(r

0

/r)

6

− 1

.

The second term is slightly tricky, so we’ll restrict to the case of high temperature,

hence small β. We can write

e

βU

0

(r

0

/r)

6

≈ 1 + βU

0

r

0

r

6

+ ···

Then we get

I ≈

4

3

πr

3

0

+

4πU

0

kT

Z

∞

r

0

dr

r

6

0

r

6

=

4πr

3

0

3

U

0

kT

− 1

.

For large

T

, this is negative, and so the repulsive interaction dominates. We can

plug this into our equation of state, and find

pV

NkT

≈ 1 −

N

V

a

kT

− b

,

where

a =

2πr

3

0

3

U

0

, b =

2πr

3

0

3

.

We can invert this to get

kT =

V

N

p +

N

2

V

2

a

1 +

N

V

b

−1

.

Taking the Taylor expansion of the inverse and truncating higher order terms,

we obtain

kT =

p +

N

2

V

2

a

V

N

− b

.

This is the van der Waals equation of state, which is valid at low density

(Nr

3

0

/V 1) and high temperature (βU

0

1).

We can also write this as

p =

NkT

V − bN

− a

N

2

V

2

.

In this form, we see that

a

gives a reduced pressure due to long distance attractive

force, while we can view the

b

contribution as saying that the atoms take up

space, and hence reduces the volume.

By why exactly the factor of

bN

? Imagine two atoms living right next to

each other.

r

0

The value

r

0

was chosen to be the distance between the two cores, so the volume

of each atom is

4π

3

r

0

2

3

,

which is not

b

. We might think the right way to do this problem is to look at

how much volume the atom excludes, i.e. the dashed volume. This is

Ω =

4

3

πr

3

0

= 2b.

This is again not right. So we probably want to think more carefully.

Suppose we have

N

particles. When we put in the first atom, the amount

of space available to put it is

V

. When we now try to put in the second, the

available space is

V −

Ω (assuming the two atoms are not too close together so

that they don’t exclude the same volume). Similarly, when we put in the third,

the available space is V − 2Ω.

Diving by

N

! for indistinguishability, we find that the total phase space

volume available for placing our particles is

1

N!

V (V − Ω)(V − 2Ω) ···(V − (N − 1)Ω) ≈

1

N!

V

N

1 −

N

2

2

Ω

V

+ ···

≈

1

N!

V −

NΩ

2

N

,

which explains why the reduced volume for each particle is Ω/2 instead of Ω.

Now suppose we want to find higher-order corrections to the partition function

in the Virial expansion. The obvious thing might be to take them to be the

contributions by

PP

f

jk

f

`m

. However, this is not quite right. If we try to keep

track of how large the terms are carefully, we will get a different answer.

The way to do it properly would be via the cluster expansion, which involves

using nice diagrams to figure out the right terms to include, similar to how

Feynman diagrams are used to do perturbation theory in quantum field theory.

Unfortunately, we do not have the time to go into the details.